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Without using trigonometric tables evaluate:
$\dfrac{{\sin {{63}^0}}}{{\cos {{27}^0}}} + \dfrac{{\cos {{39}^0}}}{{\sin {{51}^0}}} - \sin {23^0}.\sec {67^0} + {\csc ^2}{30^0}$.

Answer
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326.7k+ views
Hint: In this question apply basic trigonometric properties such as $\sin \theta = \cos \left( {{{90}^0} - \theta } \right),{\text{ cos}}\theta = \sin \left( {{{90}^0} - \theta } \right),{\text{ sec}}\theta {\text{ = }}\dfrac{1}{{\cos \theta }}$, so use these concepts to reach the solution of the question.

Given equation is
$\dfrac{{\sin {{63}^0}}}{{\cos {{27}^0}}} + \dfrac{{\cos {{39}^0}}}{{\sin {{51}^0}}} - \sin {23^0}.\sec {67^0} + {\csc ^2}{30^0}$
Now as we know that
$\sin \theta = \cos \left( {{{90}^0} - \theta } \right),{\text{ cos}}\theta = \sin \left( {{{90}^0} - \theta } \right),{\text{ sec}}\theta {\text{ = }}\dfrac{1}{{\cos \theta }}$, so use these properties in above equation we have,
$\dfrac{{\cos \left( {{{90}^0} - {{63}^0}} \right)}}{{\cos {{27}^0}}} + \dfrac{{\sin \left( {{{90}^0} - {{39}^0}} \right)}}{{\sin {{51}^0}}} - \cos \left( {{{90}^0} - {{23}^0}} \right).\dfrac{1}{{\cos {{67}^0}}} + {\csc ^2}{30^0}$
$ \Rightarrow \dfrac{{\cos \left( {{{27}^0}} \right)}}{{\cos {{27}^0}}} + \dfrac{{\sin \left( {{{51}^0}} \right)}}{{\sin {{51}^0}}} - \cos \left( {{{67}^0}} \right).\dfrac{1}{{\cos {{67}^0}}} + {\csc ^2}{30^0}$
$
   \Rightarrow 1 + 1 - 1 + {\csc ^2}{30^0} \\
   = 1 + {\csc ^2}{30^0} \\
$
Now we know that $\csc {30^0} = 2$, so substitute this value in above equation we have
$ \Rightarrow 1 + {\csc ^2}{30^0} = 1 + {\left( 2 \right)^2} = 1 + 4 = 5$
So, this is the required answer of the given equation.

Note: In such types of questions the key concept we have to remember is that always remember all the basic trigonometric properties which is stated above then simplify the given equation according to these properties and then apply the value of $\csc {30^0} = 2$, and simplify, we will get the required value of the given equation which is the required answer.
Last updated date: 24th May 2023
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